What is the value of the definite integral of x^2/(x^3+1)^2 from x = 1 to x = 2?

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Let f(x) = x^2/(x^3+1)^2

We need to find the integral of f(x) from 1 to 2.

Let F(x) = Int f(x)

==> The definite integral is:

I = F(2) = F(1).

Let us determine F(x).

==> F(x) = Int = x^2/(x^3+1)^2 dx

Let us assume that x^3+1 = u ==> du 3x^2 dx

==> F(x) = Int (1/u^2) * du/3

               = Int du/ 3u^2

                = (1/3) Int u^-2 du = (1/3) u^-1/-1 = -1/3u

==> F(x) = -1/3u

Now we will substitute back u= x^3 +1

==> F(x) = -1/3(x^3+1)

==> F(2) = -1/3(8+1) = -1/27

==> F(1) = -1/3(1+1) = -1/6

==> I = -1/27 + 1/6 = -2/54 + 9/54 = 7/54

Then the definite integral is 7/54.

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I'm sure you mean the value of the definite integral of x^2/(x^3+1)^2 from x = 1 to x = 2.

First we find the indefinite integral of x^2/(x^3+1)^2

Int [x^2/(x^3+1)^2 dx]

let u = x^3 + 1, du = 3x^2 dx or du/3 = x^2 dx

=> Int [ (1/3) * (1/ u^2) du]

=> 1/3 [ u^-2 du]

=> (1/3)(-1) u^-1 + C

substitute u with x^3 + 1

=> (-1/3)(1/ x^3 + 1) + C

For x = 2

(-1/3)(1/ x^3 + 1) + C = (-1/3)(1/9) + C

For x = 1

(-1/3)(1/ x^3 + 1) + C = (-1/3)(1/2) + C

The difference is (-1/3)(1/9) + C - (-1/3)(1/2) + C

=> (-1/3)(1/9 - (1/2)

=> (-1/3)( -7/18)

=> 7/54

The required value is 7/54

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